Question

If we have an equation as ${{x}^{2}}+{{y}^{2}}=1$, then find the differential equation.
(a) $y{y}''+{{\left( {{y}'} \right)}^{2}}+1=0$
(b) $y{y}''+2{{\left( {{y}'} \right)}^{2}}+1=0$
(c) $y{y}''-2{{\left( {{y}'} \right)}^{2}}+1=0$
(d) $y{y}''+{{\left( {{y}'} \right)}^{2}}-1=0$
(e) $y{y}''-2{{\left( {{y}'} \right)}^{2}}-1=0$

Answer 1

Hint: For solving this question, we will use some standard results of differentiation as $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ , $\dfrac{d\left( {{y}^{n}} \right)}{dx}=n{{y}^{n-1}}\dfrac{dy}{dx}$ and apply the product rule of differentiation to differentiate the given equation with respect to $x$. And we will differentiate the given equation two times and solve further to get the correct answer.

Complete step-by-step solution -
Given:
It is given that, ${{x}^{2}}+{{y}^{2}}=1$ and we have to generate a differential equation between $y$ , ${y}'$ and ${y}''$ .
Now, before we proceed we should know the following formulas and concepts of differential calculus:
1. If $y=f\left( x \right)\cdot g\left( x \right)$ , then $\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$ . This is also known as the product rule of differentiation.
2. If $y=f\left( x \right)$ , then ${y}'=\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right) \right)}{dx}$ & ${y}''=\dfrac{d\left( {{y}'} \right)}{dx}$ .
3. $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ and $\dfrac{d\left( {{y}^{n}} \right)}{dx}=n{{y}^{n-1}}\dfrac{dy}{dx}$
Now, we will use the above-mentioned formulas and concepts to differentiate ${{x}^{2}}+{{y}^{2}}=1$ with respect to $x$ . Then,
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}=1 \\
 & \Rightarrow \dfrac{d\left( {{x}^{2}}+{{y}^{2}} \right)}{dx}=\dfrac{d\left( 1 \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( 1 \right)}{dx} \\
\end{align}$
Now, as $1$ is a constant term so, we can write $\dfrac{d\left( 1 \right)}{dx}=0$ , and we will use the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ to write $\dfrac{d\left( {{x}^{2}} \right)}{dx}=2x$ , $\dfrac{d\left( {{y}^{n}} \right)}{dx}=n{{y}^{n-1}}\dfrac{dy}{dx}$ to write $\dfrac{d\left( {{y}^{2}} \right)}{dx}=2y\dfrac{dy}{dx}$ in the above equation. Then,
$\begin{align}
  & \dfrac{d\left( {{x}^{2}} \right)}{dx}+\dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( 1 \right)}{dx} \\
 & \Rightarrow 2x+2y\dfrac{dy}{dx}=0 \\
 & \Rightarrow x+y\dfrac{dy}{dx}=0 \\
\end{align}$
Now, we will write $\dfrac{dy}{dx}={y}'$ in the above equation. Then,
$\begin{align}
  & x+y\dfrac{dy}{dx}=0 \\
 & \Rightarrow x+y{y}'=0 \\
\end{align}$
Now, we will differentiate the above equation with respect to $x$ again. Then,
$\begin{align}
  & x+y{y}'=0 \\
 & \Rightarrow \dfrac{d\left( x+y{y}' \right)}{dx}=0 \\
 & \Rightarrow \dfrac{dx}{dx}+\dfrac{d\left( y{y}' \right)}{dx}=0 \\
\end{align}$
Now, we will use the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ to write $\dfrac{dx}{dx}=1$ and apply the product rule of differentiation to write $\dfrac{d\left( y{y}' \right)}{dx}={y}'\dfrac{dy}{dx}+y\dfrac{d\left( {{y}'} \right)}{dx}$ in the above equation. Then,
$\begin{align}
  & \dfrac{dx}{dx}+\dfrac{d\left( y{y}' \right)}{dx}=0 \\
 & \Rightarrow 1+{y}'\dfrac{dy}{dx}+y\dfrac{d\left( {{y}'} \right)}{dx}=0 \\
\end{align}$
Now, we will $\dfrac{dy}{dx}={y}'$ and $\dfrac{d\left( {{y}'} \right)}{dx}={y}''$ in the above equation. Then,
$\begin{align}
  & 1+{y}'\dfrac{dy}{dx}+y\dfrac{d\left( {{y}'} \right)}{dx}=0 \\
 & \Rightarrow 1+{y}'\times {y}'+y{y}''=0 \\
 & \Rightarrow y{y}''+{{\left( {{y}'} \right)}^{2}}+1=0 \\
\end{align}$
Now, from the above result, we conclude that, $y{y}''+{{\left( {{y}'} \right)}^{2}}+1=0$ .
Thus, if ${{x}^{2}}+{{y}^{2}}=1$ , then $y{y}''+{{\left( {{y}'} \right)}^{2}}+1=0$ .
Hence, (a) will be the correct option.

Note: Here, the student should know how to apply the product rule of differentiation to find the differentiation of functions of the form $y=f\left( x \right)\cdot g\left( x \right)$ . After that, while starting the solution we should be aware of the options and as in all the options term ${y}''$ is present so, we should differentiate the given equation at least twice. Moreover, though the question is very easy, we should avoid making calculation mistakes while solving and after getting the final answer we should match it correctly and select the correct option.